2A1(s) + 35n?+ (ag) -+ 2A13+ (ag) + 35n (s)
• Al loses electrons and is oxidized. Sa?* gains electrons and is reduced.
• Sn?+ gains electrons and is oxidized. Al loses electrons and is reduced.
• Sm?+ loses electrons and is oxidized. Al gains electrons and is reduced.
• Al gains electrons and is oxidized. Sn?+ loses electrons and is reduced.
The correct answer is: Sn2+ gains electrons and is oxidized. Al loses electrons and is reduced.
Based on the given chemical equation and the information provided, the correct statement is:
• Al loses electrons and is oxidized. Sn^2+ gains electrons and is reduced.
To determine which statement is correct, we need to analyze the given chemical equation and identify the elements undergoing oxidation and reduction.
Let's break down the equation:
2A1(s) + 35n?(ag) -+ 2A13+(ag) + 35n(s)
We see two elements present: Al (aluminum) and Sn (tin). In order to determine whether an element is oxidized or reduced, we need to compare the oxidation states of the elements before and after the reaction.
First, let's look at the oxidation states of Al:
In the reactant side: Al has an oxidation state of 0 because it is in its elemental form (Al(s)).
In the product side: Al has an oxidation state of +3 because it forms Al3+ ions with a charge of +3 (A13+(ag)).
Therefore, Al went from an oxidation state of 0 to +3, indicating that Al lost three electrons and was oxidized (loses electrons = oxidation).
Now, let's analyze the oxidation state of Sn:
In the reactant side: Sn has an oxidation state of +4 because it is in the form of Sn4+ ions (Sn?+(ag)).
In the product side: Sn has an oxidation state of 0 because it is in its elemental form (Sn(s)).
Therefore, Sn went from an oxidation state of +4 to 0, indicating that Sn gained four electrons and was reduced (gains electrons = reduction).
Based on the analysis, the correct statement is:
• Al loses electrons and is oxidized. Sn?+ gains electrons and is reduced.